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On the Wikipedia entry of Darcy's law, a derivation of Darcy's law from Stokes equation is provided. The derivation starts at the Stokes equation, which reads:

$$ \mu \nabla^2 u_i + \rho g_i - \partial_i p = 0 $$

where $\mu$ is the viscosity, $u$ the flow velocity, $\rho$ the fluid density, $g$ acceleration due to gravity, $p$ the fluid pressure, and $\partial$ denotes the partial derivative, all taken in the $i$-th direction ($x$, $y$, $z$, etc.). It is then said that:

Assuming the viscous resisting force is linear with the velocity we may write: $$ - \left(k_{ij} \right)^{-1} \mu \phi u_j + \rho g_i - \partial_i p = 0 $$

I fail to see how this assumption of leads to $\nabla^2 u_i = - \left(k_{ij} \right)^{-1} \phi u_j$. Would someone be so kind to explain this step in more detail?

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I don't like the derivation in the Wiki article because it leaves out all the steps. To get their equation, what you can do is assume a continuous distribution of pore number density and pore diameter with pore direction. When you integrate over these pore direction distributions, you end up with an anisotropic permeability tensor.

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  • $\begingroup$ Assuming that $k_{ij} = k I_{ij}$ (i.e. $k$ is isotropic), could you show me how $\nabla^2 u_i$ essentially reduces to $u_i$ in this case? $\endgroup$ – MPA Nov 4 '18 at 8:39
  • $\begingroup$ For a pore pointing in any arbitrary direction, the Hagen-Poiseulle equation applies within the pore (assuming constant pore diameter). The equation gives the pressure gradient in a tube in proportion to the mean velocity of flow in the tube. This is the solution to your differential equation for the tube. $\endgroup$ – Chet Miller Nov 4 '18 at 12:22

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